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A120934
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Least prime p such that the interval [p,p+log(p)] contains n primes.
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3
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2, 11, 457, 3251, 165701, 10526557, 495233351, 196039655873, 10687033762033, 79006533276941, 4313367040646743, 1740318019946551931
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OFFSET
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1,1
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COMMENTS
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Soundararajan states that, on average, there is one prime in the interval [k,k+log(k)] for any number k. Is there an upper limit to the number of primes in such an interval? Not if the prime k-tuple conjecture is true, in which case a(n) exists for all n. Note that a(n) > e^A008407(n). See A120935 for the largest prime in the interval.
a(n) begins a sequence of n primes whose prime pattern is one of the patterns in the n-th row of A186634. For example, the sequence of four consecutive primes beginning with 3251 is (3251, 3253, 3257, 3259), which has pattern (0, 2, 6, 8), which is in the 4th row of A186634.
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LINKS
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FORMULA
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This sequence grows superexponentially; a weak lower bound is a(n) >> (log n)^n. It seems that a(n) > n^n. - Charles R Greathouse IV, Apr 18 2012
A lower bound is a(n) > e^A008407(n). a(n) < b*e^A008407(n), for 2 <= n <= 12, b < 1.49. For 9 <= n <= 12, b < 1.0006. a(13) > 701673591209763173865. - Florian Baur, Jul 12 2023 [Corrected by Pontus von Brömssen, Nov 12 2023]
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EXAMPLE
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a(2)=11 because p=11 is the first prime with log(p) > 2 and 11+2 is prime.
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MATHEMATICA
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i=1; Table[While[p=Prime[i]; PrimePi[p+Log[p]]-PrimePi[p]+1< n, i++ ]; p, {n, 5}]
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PROG
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(PARI)
my(v = vector(n, k, prime(k)), i = 1);
while(v[(i - 2) % n + 1] - v[i] > floor(log(v[i])),
v[i] = nextprime(v[(i - 2) % n + 1] + 1);
i = i % n + 1; );
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CROSSREFS
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KEYWORD
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hard,more,nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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